package org.usmile.algorithms.leetcode.middle;

/**
 * 529. 扫雷游戏
 *
 * 让我们一起来玩扫雷游戏！
 * 给你一个大小为 m x n 二维字符矩阵 board ，表示扫雷游戏的盘面，其中：
 * 'M' 代表一个 未挖出的 地雷，
 * 'E' 代表一个 未挖出的 空方块，
 * 'B' 代表没有相邻（上，下，左，右，和所有4个对角线）地雷的 已挖出的 空白方块，
 * 数字（'1' 到 '8'）表示有多少地雷与这块 已挖出的 方块相邻，
 * 'X' 则表示一个 已挖出的 地雷。
 * 给你一个整数数组 click ，其中 click = [clickr, clickc] 表示在所有 未挖出的 方块（'M' 或者 'E'）中的下一个点击位置（clickr 是行下标，clickc 是列下标）。
 * 根据以下规则，返回相应位置被点击后对应的盘面：
 * 如果一个地雷（'M'）被挖出，游戏就结束了- 把它改为 'X' 。
 * 如果一个 没有相邻地雷 的空方块（'E'）被挖出，修改它为（'B'），并且所有和其相邻的 未挖出 方块都应该被递归地揭露。
 * 如果一个 至少与一个地雷相邻 的空方块（'E'）被挖出，修改它为数字（'1' 到 '8' ），表示相邻地雷的数量。
 * 如果在此次点击中，若无更多方块可被揭露，则返回盘面。
 *
 * 示例 1：
 * 输入：board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
 * 输出：[["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
 *
 * 示例 2：
 * 输入：board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
 * 输出：[["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
 *
 * 提示：
 * m == board.length
 * n == board[i].length
 * 1 <= m, n <= 50
 * board[i][j] 为 'M'、'E'、'B' 或数字 '1' 到 '8' 中的一个
 * click.length == 2
 * 0 <= clickr < m
 * 0 <= clickc < n
 * board[clickr][clickc] 为 'M' 或 'E'
 */
public class _0529 {
}

class _0529_Solution {
    char[][] board;
    int rows;
    int cols;
    boolean[][] visited;
    int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};

    public char[][] updateBoard(char[][] board, int[] click) {
        this.board = board;
        this.rows = board.length;
        this.cols = board[0].length;
        this.visited = new boolean[rows][cols];

        if (board[click[0]][click[1]] == 'M') {
            board[click[0]][click[1]] = 'X';
        } else {
            dfs(click[0], click[1]);
        }

        return board;
    }

    private void dfs(int row, int col) {
        if (!inArea(row, col) || board[row][col] != 'E' || visited[row][col])
            return;

        visited[row][col] = true;

        int mines = 0;
        for (int[] dir : dirs) {
            int nextRow = row + dir[0];
            int nextCol = col + dir[1];
            if (inArea(nextRow, nextCol) && board[nextRow][nextCol] == 'M') {
                mines++;
            }
        }

        if (mines > 0) {
            board[row][col] = (char) (mines + '0');
        } else {
            board[row][col] = 'B';
            for (int[] dir : dirs) {
                int nextRow = row + dir[0];
                int nextCol = col + dir[1];
                dfs(nextRow, nextCol);
            }
        }
    }

    private boolean inArea(int row, int col) {
        return row >= 0 && row < rows && col >= 0 && col < cols;
    }
}
